sin a 4 5 cos b 5 13

DiketahuiSin A= 𝟒/( 𝟓) dan Sin B= 𝟓/( 𝟏𝟑)(A dan B sudut lancip). Nilai Cos (A ─ B)persamaan trigonometri yang termasuk identitas adalahpersamaan trigon Answer(1 of 5): The answer is: \cos \left ( A+B \right )= \frac{45\sqrt{17}-32}{221} follow the steps below: Iniadalah salah satu sifat dari sinus dan cosinus. Dari rumus 1, kita bisa mencari cos a. Mencari "cos a" Langsung saja gunakan rumus diatas.. sin²a + cos²a = 1 Diketahui : sin a = x (x)² + cos²a = 1. x² + cos²a = 1. pindahkan x² ke ruas kanan menjadi -x² Ifthe acute angle θ is given, then any right triangles that have an angle of θ are similar to each other. This means that the ratio of any two side lengths depends only on θ.Thus these six ratios define six functions of θ, which are the trigonometric functions.In the following definitions, the hypotenuse is the length of the side opposite the right angle, opposite represents the side SinaSinb is the trigonometry identity for two different angles whose sum and difference are known. It is applied when either the two angles a and b are known or when the sum and difference of angles are known. It can be derived using angle sum and difference identities of the cosine function cos (a + b) and cos (a - b) trigonometry identities which are some of the Lieu De Rencontre Fontenay Le Comte. We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169⇒cosA=35 and sinB=1213 Now, sinA+B=sinA cosB+cosA sinB =45×513+35×1213=2065+3665=20+3665=5665 ii We have, sinA=45 and B=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1452 and sin B=√1−5132 ⇒cosA=√1−1615 sin B=√1−25169 ⇒cosA=√25−1625 and sin B=√169−25169 ⇒cosA=√925 and sin B=√144169 cosA=35 and sinB=1213 Now, cosA+B=cosA cosB−sinA sinB =35×513−45×1213 =1565−4865 =15−4865=−3365 iii We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 and sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=√169−25169 ⇒cosA=√925 and sinB=√144169 ⇒cosA=35 and sinB=1213 Now, sinA−B=sinA cosB−cosA sinB =45×513−35×1213 =2065+3665=20−3665=−1665 iv We have, sinA=45 and cosB=513 ∴cosA=√1−sin2A and sinB=√1−cos2B ⇒cosA=√1−452 sinB=√1−5132 ⇒cosA=√1−1625 and sinB=√1−25169 ⇒cosA=√25−1625 and sinB=169−25169 ⇒cosA=√925 and sinB=1213 Now, cosA−B=cosA cosB+sinA sinB =35×513+45×1213 =1565+4865 =15+4865 =6365

sin a 4 5 cos b 5 13